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21x^2+20x+4=0.5
We move all terms to the left:
21x^2+20x+4-(0.5)=0
We add all the numbers together, and all the variables
21x^2+20x+3.5=0
a = 21; b = 20; c = +3.5;
Δ = b2-4ac
Δ = 202-4·21·3.5
Δ = 106
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-\sqrt{106}}{2*21}=\frac{-20-\sqrt{106}}{42} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+\sqrt{106}}{2*21}=\frac{-20+\sqrt{106}}{42} $
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